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# Stuck on C3 question… watch

1. This question (part ii) is really annoying me!! Because I almost have the answer but can’t get to the final one and seem to have done everything right…

I started off with y=mx+c, found xsin2x = 0 at 0,π& 2π, clearly the graph shows P is the first root after 0 so it’s π, then I subbed 2x=π (x=π/2) into tan2x + 2x to get my gradient which again was π, then I wrote that in the form of y=π(x) + c, subbed in P(π/2,0) and got y=π(x)-(π^2/2), so c=π^2/2 when I multiply the equation by 2 I get 2y=2πx-π^2… but they want it in the form of 2πx + 2y = π^2, I cant rearrange my answer into that form…

I’d really appreciate it if someone could show me where I’m going wrong, It could just be a sign error somewhere in my working
2. (Original post by Zxyn)

This question (part ii) is really annoying me!! Because I almost have the answer but can’t get to the final one and seem to have done everything right…

I started off with y=mx+c, found xsin2x = 0 at 0,π& 2π, clearly the graph shows P is the first root after 0 so it’s π, then I subbed 2x=π (x=π/2) into tan2x + 2x to get my gradient which again was π, then I wrote that in the form of y=π(x) + c, subbed in P(π/2,0) and got y=π(x)-(π^2/2), so c=π^2/2 when I multiply the equation by 2 I get 2y=2πx-π^2… but they want it in the form of 2πx + 2y = π^2, I cant rearrange my answer into that form…

I’d really appreciate it if someone could show me where I’m going wrong, It could just be a sign error somewhere in my working
isn't the gradient function, it's just an equation which is satisfied when you plug in the coordinates of the turning points.
3. (Original post by RDKGames)
isn't the gradient function, it's just an equation which is satisfied when you plug in the coordinates of the turning points.
Ohhhhh thanks!!! I tried it with the gradient function (sin2x + 2xcos2x) and got -π which fixes everything!

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